Mean And Variance Polarization

Cauchy Distribution does not have a mean, but

$\int^\infty_0{x.f(x)}dx=\int^\infty_0{\cfrac{2x^2}{\pi(1+x^4)}}dx=\cfrac{1}{\sqrt{2}}$

when we consider both $-\theta$ and $\theta$

$E\left\{x\right\}=2\cfrac{1}{\sqrt{2}}=\sqrt{2}$

The emitted photons, assuming that $\theta$ is uniformly distributed, $-\infty\lt\theta\lt\infty$, has an expected value of $\sqrt{2}$.

But,

$\int{x^2.f(x)}dx=\int{\cfrac{2x^3}{\pi(1+x^4)}}dx=\cfrac{1}{2\pi}ln(x^4+1)$

does not converge for all values of $\theta$, however for the range $0\le x\le \cfrac{\pi}{2}$,

$E\left\{x^2\right\}=0.31169$

in which case, over the same range,

$E\left\{x\right\}=0.314044$

and the variance is,

$Var\left\{x\right\}=2*0.31169-(2*0.314044)^2$

$Var\left\{x\right\}=0.22889$,   $\sigma=\sqrt{Var\left\{x\right\}}=0.4784$

Polarization mean and variance has little meaning, compared to mode at $\cfrac{1}{\sqrt[4]{3}}$.